Problem 4
Question
Use (10.1.1) to determine \(L[f]\). \(f(t)=\sin b t,\) where \(b\) is constant.
Step-by-Step Solution
Verified Answer
The Laplace transform of the function \(f(t) = \sin(bt)\) is \(L[f(t)] = L[\sin(bt)] = \frac{b^2}{s^2 + b^2}\).
1Step 1: Set Up the Integration Formula
First, let's substitute the given function into the Laplace transform formula:
\[F(s) = L[\sin(bt)] = \int_{0}^{\infty} e^{-st} \sin(bt) dt\]
Now we have the integral that we need to evaluate.
2Step 2: Integration by Parts
To evaluate the integral, we will use integration by parts. Let's choose our functions as follows:
\[u = e^{-st}, \quad dv = \sin(bt) dt\]
Then, we can differentiate u and integrate dv:
\[du = -se^{-st} dt, \quad v = -\frac{1}{b}\cos(bt)\]
Now we can apply integration by parts:
\[F(s) = L[\sin(bt)] = -\frac{1}{b}\int_{0}^{\infty} e^{-st}\cos(bt) dt + \frac{s}{b}\int_{0}^{\infty} e^{-st}\sin(bt) dt\]
We'll use integration by parts again, this time choosing:
\[u = e^{-st}, \quad dv = \cos(bt) dt\]
Which gives us:
\[du = -se^{-st} dt, \quad v = \frac{1}{b}\sin(bt)\]
Applying integration by parts a second time, we get:
\[F(s) = L[\sin(bt)] = -\frac{1}{b}\left[\frac{1}{b}\sin(bt)e^{-st}\bigg|_{0}^{\infty} + \frac{s}{b}\int_{0}^{\infty} e^{-st}\sin(bt) dt\right] - \frac{s}{b}\left[\frac{1}{b}\cos(bt)e^{-st}\bigg|_{0}^{\infty} + \frac{s}{b}\int_{0}^{\infty} e^{-st}\cos(bt) dt\right]\]
3Step 3: Evaluate the Limits and Simplify
Now, we'll evaluate the limits at the vertical bars and simplify:
\[F(s) = L[\sin(bt)] = \frac{s}{b^2}L[\sin(bt)] - \frac{s^2}{b^3}L[\cos(bt)]\]
If we assume that \(s > 0\), the limits of \(e^{-st}\sin(bt)\) and \(e^{-st}\cos(bt)\) as \(t \rightarrow \infty\) will be zero.
Solving for \(L[\sin(bt)]\), we get:
\[F(s) \left(1 - \frac{s^2}{b^3}\right) = \frac{s}{b^2}F(s)\]
Now divide both sides by \(\left(1 - \frac{s^2}{b^3}\right)\) and \(\frac{s}{b^2}\):
\[F(s) = L[\sin(bt)] = \frac{b^2}{s^2 + b^2}\]
Thus, the Laplace transform of the function \(f(t) = \sin(bt)\) is:
\[L[f(t)] = L[\sin(bt)] = \frac{b^2}{s^2 + b^2}\]
Key Concepts
Laplace transformIntegration by partsDifferential equations
Laplace transform
The Laplace transform is a powerful mathematical tool used to simplify the analysis of linear time-invariant systems, particularly in the fields of engineering and physics. It transforms a time-domain function into a complex frequency-domain representation, which can make solving differential equations more manageable.
For a given function \(f(t)\), the Laplace transform, denoted by \(L[f(t)]\), is defined as:
\[ L[f(t)] = \int_{0}^{\infty} e^{-st}f(t) dt \]
where \(s\) is a complex number, and \(e^{-st}\) is the decaying exponential that weights the function. For the sine function \(f(t) = \sin(bt)\), the transform provides a way to convert this oscillating function into an algebraic form that is easier to work with in the frequency domain. This simplification is instrumental in solving differential equations with sinusoidal functions or analyzing systems that respond to such functions.
The Laplace transform also has properties that make it particularly appealing for solving linear equations, such as linearity, differentiation, and integration. Using these properties and the transform pairs found in Laplace transform tables, many complex time-domain problems are reduced to simple algebra.
For a given function \(f(t)\), the Laplace transform, denoted by \(L[f(t)]\), is defined as:
\[ L[f(t)] = \int_{0}^{\infty} e^{-st}f(t) dt \]
where \(s\) is a complex number, and \(e^{-st}\) is the decaying exponential that weights the function. For the sine function \(f(t) = \sin(bt)\), the transform provides a way to convert this oscillating function into an algebraic form that is easier to work with in the frequency domain. This simplification is instrumental in solving differential equations with sinusoidal functions or analyzing systems that respond to such functions.
The Laplace transform also has properties that make it particularly appealing for solving linear equations, such as linearity, differentiation, and integration. Using these properties and the transform pairs found in Laplace transform tables, many complex time-domain problems are reduced to simple algebra.
Integration by parts
Integration by parts is a technique derived from the product rule of differentiation and is used to integrate products of functions. It is particularly useful when dealing with products where one function is easily integrable, and the other is easily differentiable. This technique is represented by the formula:
\[ \int u dv = uv - \int v du \]
In the context of finding the Laplace transform of the sine function, integration by parts is used twice because direct integration isn’t straightforward due to the infinite limits of integration and the oscillatory nature of the sine function.
The choice of \(u\) and \(dv\) is crucial for simplifying the integration process. We generally select \(u\) to be a function that becomes simpler when differentiated, and \(dv\) to be a function that does not become more complicated when integrated. In the solution provided, integration by parts helps to break down the complex integral into simpler parts, eventually leading to an expression for the Laplace transform that can be easily handled.
\[ \int u dv = uv - \int v du \]
In the context of finding the Laplace transform of the sine function, integration by parts is used twice because direct integration isn’t straightforward due to the infinite limits of integration and the oscillatory nature of the sine function.
The choice of \(u\) and \(dv\) is crucial for simplifying the integration process. We generally select \(u\) to be a function that becomes simpler when differentiated, and \(dv\) to be a function that does not become more complicated when integrated. In the solution provided, integration by parts helps to break down the complex integral into simpler parts, eventually leading to an expression for the Laplace transform that can be easily handled.
Differential equations
Differential equations are equations that involve derivatives of one or more variables. In physical systems, they are used to describe the rate of change of a system and are fundamental in expressing the dynamics of systems in engineering, physics, and other sciences.
Linear differential equations with constant coefficients can often be solved using the Laplace transform. By taking the Laplace transform of both sides of a differential equation, one can often transform a differential equation in the time domain into an algebraic equation in the frequency domain. This process simplifies the solution of the equation because algebraic equations are typically less complex to solve than differential equations.
Once the Laplace transform of the function is found, one then employs the inverse Laplace transform to revert back to the time domain and find the general solution to the original differential equation. The exercise on finding the Laplace transform of the sine function exemplifies how the Laplace transform facilitates the process of solving differential equations that have sinusoidal inputs or inherent oscillatory aspects.
Linear differential equations with constant coefficients can often be solved using the Laplace transform. By taking the Laplace transform of both sides of a differential equation, one can often transform a differential equation in the time domain into an algebraic equation in the frequency domain. This process simplifies the solution of the equation because algebraic equations are typically less complex to solve than differential equations.
Once the Laplace transform of the function is found, one then employs the inverse Laplace transform to revert back to the time domain and find the general solution to the original differential equation. The exercise on finding the Laplace transform of the sine function exemplifies how the Laplace transform facilitates the process of solving differential equations that have sinusoidal inputs or inherent oscillatory aspects.
Other exercises in this chapter
Problem 4
Make a sketch of the given function on the interval \([0, \infty)\). $$f(t)=t\left(1-u_{1}(t)\right)$$.
View solution Problem 4
Determine the Laplace transform of the given function \(f.\) $$f(t)=\sin (t-\pi / 4) u_{\pi / 4}(t)$$.
View solution Problem 4
Use the Laplace transform to solve the given initial-value problem. \(y^{\prime}+2 y=4 t, \quad y(0)=1\).
View solution Problem 4
Show that the given function is of exponential order. $$f(t)=e^{3 t} \sin 4 t.$$
View solution